1 




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UBRARY OF CONGRESS 



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Class __7M>C)"t 
Bookx/4f 



COPYRIGHT DEPOSIT. 



NOTES AND PROBLEMS 



TO ACCOMPANY AN 



INTEODUCTORY COUESE 



IN THE 



RESISTANCE OF MATERIALS 



GIVEN IN HARVABD UNIVERSITY UNDER THE NAME OF 



ENGINEERING M 



BY 



LEWIS JEROME JOHI^SON, A.B., C.E. 

ASSISTANT PROFESSOR Or CIVIL ENGINEERING IN HARVARD UNIVERSITY 




CAMBRIDGE 

lC>ublisbe& b^ IbarvarD IHniverslt^ 

1903 



THE LIBRARY OF 
CONGRESS, 


Two Copies 


deceived 


MAR IT 


1903 


Copyughl 


Entry 


CLASS a^ 


'3 -"2 

XXe, No. 


COPY 


5 

B. 



Copyright, 1903 
By L. J. Johnson 






NOTES AND PROBLEMS 



roR 



ENGINEERING 5^ 



1. Definitions. Stress is the tendency to distortion or rupture in a body, due to the 
action of external forces. Its value is expressed numerically by the common magnitude 
of two equal and opposite forces or couples to which the external forces may be reduced. 
Read carefully Statics, Chap. IX, especially §§ 59, 61, where the four fundamental kinds of 
stress, normal sti-ess (including tension and compression), shear, flexure, and torsion, are 
defined and explained. 

Unit-Stress is the stress existing within a unit of area, the stress being understood to 
be uniform in intensity throughout that area. The phrase unit-stress at a given point is 
used to designate the stress that would be brought to bear upon a unit area if the stress- 
intensity were the same on the whole of that area as it is at the point. Unit-stress 
thus bears the same relation to a force or stress as velocity does to distance. A speed 
of a mile a minute and unit-stress at a given point of 1,000 lbs. per sq. in. are precisely 
analogous terms. The unit-stress implies no more as to the magnitude of the total stress 
than the given rate of speed does to the total distance traversed. They are alike 
merely rates. 

Computed unit-stresses, upon comparison with unit-stresses known by experience to be 
permissible for the given material under the given conditions, furnish the criterion by 
which is judged the ability of the body to render service expected of it. 

Deformation (or Strain) is the change of shape or size of the body due to a stress. 

Unit-Deformation is the amount of deformation per unit of some dimension of a body. 
For elongations and shortenings, the only deformations commonly dealt with from this 
point of view, the dimension concerned is the length of the bar or lamina subject to the 
stress. 

2. Relation of Resistance of Materials to Statics. Resistance of Materials treats of 
the behavior of materials under stress with special reference to throwing light upon 
structural design. It is the science by the aid of which one provides sufficient stiffness 
and strength in a projected structure, and estimates the margin of safety in an existing 
structure. 



In all this work a knowledge of stresses is of fundamental importance. The determi- 
nation of the stresses is, as has been seen, a matter of statics, — usually, in fact, a matter 
of pure statics, in which no thought is given to possible deformation or injury of the body 
as a result of the stresses. This assumption of rigidity simplifies the work, and, as pointed 
out in Statics, § 4, produces no error of practical importance. 

The study of strength involves the further necessity of tracing the relation between the 
unit-stresses in the various parts of a section and the total stress in the section. This 
is in effect a question of resolving a force into an infinite number of components, — a 
problem indeterminate upon the rigid body hypothesis. The actual lack of rigidity is 
therefore brought into account at this point, and by careful consideration of the nature 
of the deformations attending the various kinds of stress, the conditions required in addi- 
tion to those of pure statics are deduced. The problem is worked out accordingly, and 
the desired unit-stresses determined. 

A very large part of the science of resistance of materials is therefore strictly statical 
in its nature ; it is a study of equilibrium, using the conditions of equilibrium of a rigid 
body so far as they will suffice, and reinforcing them at points otherwise impassable by 
conditions based upon empirical knowledge of actual non-rigid bodies. 

3. Fundamental Laws. The empirical knowledge referred to at the close of the last 
section may conveniently be stated in the form of eight propositions, which form the 
basis of the science of resistance of materials so far as it is distino^uishable from statics 
as ordinarily understood. These eight propositions are as follows: — * 

A. When a small sti^ess is applied to a body a small deformation is produced, 
and on the rem,oval of the stress the body springs back to its oinginal form. For 
small stresses, then, inatericds may be regarded as perfectly elastic. 

B. Under small stresses the deformations are approximately proportional to the 
forces, or stresses, which produce them, and also approximately proportional to 
the length of the bar or body. 

The proportionality between small stresses and the accompanying deformations was 
first pointed out by Robert Hooke in 1678, though discovered by him 18 years earlier, 
and kept secret in hope of patenting some applications of the principle to manufacture of 
watch and clock springs. He had ventured to publish it in 1676 in the form of an 
anagram ceiiinosssttuv or ut tensio sic vis. The law is still known as Hooka's Law, 
and will be so spoken of in this course. 

The ratio of unit-stress to unit-deformation which Hooke observed to be constant for 
small stresses is a highly important property of a material and is given the name modu- 
lus of elasticity, or £J.-\ If /= unit-stress and s the unit-deformation, 

s 

* The statement of these principles is substantially that of Professor Merriinan in liis Mechanics of Materials. 
t This E is the so-called Young's Modulus. It is usually associated with tension, but holds also for engineering 
materials in compression. Cf. Ewing : Strength of Materials, p. 31. 



Further, if P is the stress, A the area of the section, I the length of the bar or 
lamina concerned, and A the total deformation, then f =. — and ^ = r and 

JS is the measure of the stiffness of the material and is a unit-stress, — in fact, that 
unit-stress which would produce a deformation equal to the original length of the bar, 
if this could be produced without overstepping the elastic limit (see below) or producing 
fracture, 

C. As the stress increases, there is a more or less definite point reached heyond 
which the increase in deformation is greater relatively than the increase in stress, 
and the deformations produced will not disappear entirely on the removal of 
the stress. 

The intensity of stress or unit-stress at which this change is noted is called the elastic 
limit, and the lasting deformations resulting from overstepping the elastic limit is called 
the set, or sometimes permanent set. 

The term small stress used in A and JJ means simply stress within the elastic limit, 
and since the elastic limit is not overstepped in any properly designed structure, such a 
structure is affected only by ' ' small stresses " within the meaning of A and J5, and laws 
A and £ are actually laws pertaining to the structures in common service. 

D. If the stress continues to increase, deformation will continue to increase, and 
finally the body ruptures, or '•^ fails,'' i. e. becomes so changed as to be wholly unsuited 
to voider service such as it rendered tender the smaller stress. 

For example, a tension rod or rope may be pulled entirely in two by an overload, 
while a short column of soft steel by flattening out under an ovei'load would in that way 
cease to furnish a point of support at the original and required height. A wooden joist 
may be broken entirely in two by a heavy centre load, while a steel beam would become 
useless or unserviceable by bending into awkward shape — without necessarily actually 
breaking. Such and all other forms of failure are the result of an indefinitely increas- 
ing stress. The stress under which a body fails is called the ultimate strength of the 
body. The stress per unit of area (unit-stress) at the time of failure is called the ultimate 
unit-streng-th, or for short, but somewhat loosely, simply ultimate strength. 

E. A sudden stress or shock is more injurious than a steady stress or a stress 
gradually produced. 

F. The horizontal fibres on the conmex side of a beam subject to bending are 
elongated, and those on the concave side are shortened, while somewhere between the 
tioo sides is a neutral surface which is unchanged in length. 

G. The amount of lengthening or shortening of any fibre of a moderately bent 
beam can reasonably and scfely be assumed to be directly 2^^opo7'tional to the distance 



from the neutral surface, or, in other words, a heam cross-section which is 2>lcbne 
before bending can reasonably and safely be assumed to be plane after bending, 
though inclined to its original j^osition. 

H. The unit-stresses at given 'points in the cross-section of a beam may be assumed 
to be directly proportional to the distances of these points from, the neutral surface, 
p)rovided the bending is moderate in amount, i. e. provided the elastic limit 
of the material has not been overstepped at any point of the section. 

Law H, though a direct deduction from Laws B and G, is here given this separate 
enunciation on account of its great importance. 

4. Fundamental Relation between Stresses and Unit-Stresses One of Pure Statics. 
The connecting link between stresses and unit-stresses is the principle of pure statics 
that if all segments of a body or portions of a body are to remain in equilibrium in spite 
of forces applied to the body, the internal resistance at the boundary between any two 
segments must be as great as the total external effort on either segment, or, in other words, 

The internal forces at the end of a segment must balance all the external forces 
to which the seg7nent is subject. 

5. Stress Expressible in Terms of Unit-Stress Multiplied by an Area, or by an 
Area and a Length. The two simpler kinds of stress, normal stress (tension and com- 
pression) and shear, are measured by the magnitudes of single forces and are hence 
expressed numerically in weight-units such as pounds, while unit-stresses are of coui'se 
expressed in weight-units per unit of area, as, for example, in pounds per square inch. In 
order then to express any force or normal stress or shear in terms of unit-stress, the 
unit-stress must be multiplied by an area. A couple, being expressible by a force multi- 
plied by a length, can accordingly be expressed in terms of a unit-stress times an area 
times a length. That is 

Tension or compression, and shear can be equated with a unit-stress vnultiplied 
by an area, and flexure and torsion with a unit-stress times an area times a length. 

In the next four sections this principle will be applied together with those of the 
two preceding sections, and the relations between the four kinds of stress, normal stress, 
shear, flexure, and torsion, and their corresponding unit-stresses, will be expressed in 
algebraic form. 

6. Tension and Compression. By observation of the behavior of a bar subject to 
tension from a fair pull, or of a block subject to compression from a fair central load 
and too short to be in danger of bending, it appears that the body simply lengthens or 
shortens as a whole, that is, that any plane section of the body simply moves to a position 
parallel to its original position. This means that the deformation is uniform at all points 
of the section, and it follows by Hooke's Law that the stress is uniform over the section, 
and that the unit-stress is constant throughout the section. 



For example, if a bar 4 square inches in cross-section is subject to 4,000 lbs. tension, 
the tensile unit-stress (or, more briefl}^ the tensile stress) throughout a right section of 
the bar is 1,000 lbs. per sq. in. Conversely, if the tensile stress is 1,000 lbs. per sq. in. 
in a bar 4 square inches in cross-section, the total tension in the bar is 4,000 lbs. 
The same relations between the total stress and the unit-stress would have held if the 
stress had been compression. To express these relations in general form, P may be 
taken to indicate the total stress, and/'^,^^' '^^^ -4 the tensile unit-stress, the compres- 
sion unit-stress, and the area of the section respectively. Then since P is the total nor- 
mal effort upon the section, and /"^^ and /c^ are the total normal resistances, and since 
the total resistance must equal the total effort (§4), it follows that 



P = J\A la 

P=:f,A for short blocks lb 



}<.) 



7. Shear. As to the behavior of a bar subject to shear, mathematical analysis and 
experiment make it evident that resistance to shear is not always furnished uniformly by 
all parts of a section. In some common cases, however, this variation must be very slight 
indeed, and in most cases, in practice, the shear can be treated satisfactorily l)y considering 
it uniforml}^ distributed over a definite part if not the whole of the section exposed. 

Calling, then, the unit-stress in shear's, and calling ^, the area of the portion of the 
section in which the resistance to shear may be taken as uniform, then f.,Ag will be the 
total resistance to shear at the section. 

Then if V be the shearing effort upon the segment, the effort and resistance being 
necessarily equal, there follows an equation of the same form as those for tension and 
compression above. 

F=./:,A (2) 

In rectangular sections the maximum shearing unit-stress is 50 per cent, greater than 
the mean. Equation (2), for rectangular beams, may accordingly be written P =^ ^f,^A, 
where A is the section of the beam, and /^ is as before. In I-beams and plate girders 
the shear comes to bear almost wholly on the webs and should be so assumed. A, for 
use in (2) in such cases, is assumed to be the cross-section of the web. 

This last paragraph will be developed in the lectures. 

8. Flexure. In a section subject to flexure, the variation in unit-stress from point to 
point is far too great to be overlooked. The statical principle of § 4, however, together 
with law H furnish the necessary conditions for determining the stress-intensity at any 
point of the section in terms of given quantities. 

By § 4, it may be written that 

M=so,/;?/ (3) 

Avhere 31 is the common magnitude of the two equal and opposite couples to which the 
external forces on each side of the section are reducible — i.e. M is the Hexure at the 
section — and a is the area of an infinitely small portion of the section on which the unit- 
stress has the constant -value fy and which is distant y from some convenient axis of 



8 

moments. Both the bending and resisting couples will of course, like any other couples, 
have values independent of the position of the axis so long as the axis is perpendicular 
to their plane. (Cf. Statics, p. 16.) The most convenient axis for use is, however, one 
coincident with the neutral axis of the section, and no other need ever be chosen in 
practice. The reasons for preferring the axis to be so chosen will appear as the work 
proceeds. 

In equation (3) it should be noted that afy is a force and, y being a length, afyy is 
the moment of one of the elementary forces into which the resisting couple is to be 
decomposed about the axis from which y is measured, which, as just stated, is most 
conveniently an axis coincident with the neutral axis. The position of this axis will be 
determined presently. 

Law H expressed in algebraic language becomes simply 

fi —. y 

ft Vx 

f. = iy (4) 

Where fy and yi are unit-stresses at points distant y and y^ from the neutral axis, sub- 
stituting this value of j^in (3), there results 

or, — being constant by law H, and the subscripts being no longer necessary, and, substi- 
tuting the single letter /for ^ay^, a constant. 



y 

My 



M=^- I 
y 



f=f (5) 

Hence it appears that, M and / being constant in any given problem, f will be greatest 
at the portions of the section for which y is greatest, i. e. at the points of the section most 
remote from the neutral axis, and the other extreme value of/" will be found at the point 
most remote from the neutral axis on the other side of the neutral axis. The letter c is 
commonly assigned to the absolute maximum value of y in a cross-section, and the value of 
the stress in the extreme fibre — the maximum unit-stress in flexure — may hence be written 

/' = f (6) 

For the stress at the opposite edge, the ordinate of that edge, — Cj, need only be substituted 
for c and there results 

/" = - ^ . . . (6^) 

the minus sign indicating simply that /' and /" are necessarily opposite in character. The 
extreme unit-stress in the section will be tension or compression according as the convex 
or the concave side is the more remote from the neutral axis. 



The quantity /(= Say^) in (6) is a purely geometrical property of the cross-section 
of the beam, still to be determined (in the problems bound herewith) for the sections 
commonly used for beams, such as i"ectangles, /-sections, etc. It is given the name 
moment of inertia, which should be understood to be merely an arbitrary name, thi-owing 
no more light on the real meaning of the quantity than the letter / itself. / is evidently 
a quantity of the fourth degree and is expressed in biquadratic units of leng-th, as biquadratic 
inches, or biquadratic feet. 

The length c remains to be determined before (6) can be used. This can be done by the 
aid of the statical principle of § 4. That is, the forces afy amounting as a whole to a 
couple must have zero for their algebraic sum. With the help of equation (4) this can 
be expressed algebraically 

Sa/, = Sa ^ 2/ = ^ Sa2/ = 

or, since — is a constant, %ay = 0. 

This condition holds, however, only when the y's are measured from an axis through 
the centre of gravity of a section. It follows, therefore, that for pure* flexure the neutral 
axis must pass through the centre of gravity of the section. 

The quantity c may then be defined as the ordinate, measured from an axis through 
the centre of gravity parallel to the neutral axis, of the point in the section most remote 
from such centre of gravity axis. In pure flexure c is simply the distance of the extreme 
fibre, so-called, from the neutral axis. 

Writing equation (G) in a form uniform with the corresponding equations for the 
stresses previously treated, ^. e. setting the value of the bending couple equal to the 
existing couple, it becomes 

^ = /^ (7) 

The right-hand member is usually looked upon as consisting of two quantities, f 
and - , the former being the unit-stress at the extreme fibre, and the latter a geometrical 
property of the section known as the section modulus of the section. The f takes 
account of the strength of the material of the beam, and the - takes account of the size 
and shape of its cross-section. Thus the equation and everyday experience are in harmony 
as to what contributes to the ability of a beam to resist flexure. 

Another useful check on the equation will be to see how f- is actually the moment 
of a couple — a force-magnitude times a length — as it obviously must be to be capable 
of being set equal to M. This appears on noting that / (:= '%ay'^) is the fourth power of 
a length ; that, c being a length, - is the third power of a length and capable of being- 
regarded as some area times some length ; that f being a unit-stress, multiplied by this 
area, yields a force-magnitude, and this times the length yields the moment of a couple 
as it should. 

In flexure it is evident that only a small portion of the material, that at the edge 
* That is, when the resultant of the forces on either side of the section has no component normal to the section. 



10 

most; remote from neutral axis, will render service up to its full capacity. Hence flexure 
is a kind of stress relatively costly to resist, and to be avoided where possible. 

Although, by hypothesis, (7) is applicable only within the elastic limit, nevertheless 
beams are broken by flexure and the value of/ at rupture figured by (7) just as if (7) would 
apply in such a case. This is done for want of a more rational procedure. The special 
value of / so figured is called the modulus of rupture, and is used as the ultimate value 
of/ in flexure. If (7) were strictly valid beyond the elastic limit, the modulus of rupture 
would be simpl}^ the ultimate tensile or compressive strength, but as it is, it of course 
diflfers more or less from them, as will be seen. Fortunately, experience has taught us 
values of/ which can safely be permitted for the materials commonly subjected to flexure. 
This value is of course within the elastic limit and yields valid results in (7). The irration- 
ally obtained modulus of rupture is used as a means of stating roughly the ultimate flexural 
strength of materials. 

It may be interesting to note that law H which makes possible the deduction of this 
equation dates from a time so recent as 18.'57. It is due to the French investigator, Saint- 
Venant, whom J. B. Johnson described in Modern Framed tStrnctures as the "engineer, 
physicist, and teacher who has done more than any other one to bring theory and prac- 
tice into harmony and to put both on a thoroughly scientific basis, so far as the strength 
and elasticity of engineering materials are concerned." For further account of the search 
for a satisfactory treatment of flexure, see pp. 119, 120 of the work just quoted. 

It must be pointed out in closing this discussion that equation (7) is not universally 
valid. It is valid (subject to limitations above stated) for the overwhelming majority of 
the cases of flexure which arise in practice, cases in which the plane of the bending 
couple includes an axis of symmetry of the beam section or a normal to such an axis. 
In such or other cases in which the neutral axis is normal to the plane of the bending 
couple, the equation holds, but under no other circumstances. For a fuller discussion of 
this subject see an article in the Journal of the Associated Engineering /Societies for May, 
1902, entitled The Determination of Unit-Stresses in the General Case of Flexure. 

9. Torsion. Pure torsion or twisting involves shear of varying intensity over a cross- 
section of a bar, just as flexure involves normal stress of varying intensities in various parts 
of the section. Within the elastic limit and for round bars the torsional unit-stress at any 
point is taken to be proportional to the distance of the point from the axis of the bar normal 
to the section, just as the flexural unit-stress at a point is taken to be proportional to the 
distance of the point from an axis included in the section. For bars other than round ones 
this proportionality does not hold, and the investigation of torsional unit-stresses in them 
becomes a very doubtful problem. Fortunately, round bars can almost always be selected 
for this sort of service. 

The investigation proceeds in a manner exactly analogous to that used in the case of 
flexure. In like manner then, calling Pp the twisting moment, z the distance of any 
element a of the section from the axis of the shaft, there results 

^ z 



11 

The yj will manifestly be greatest at the surface of the shaft, where z has the value ?•, and 
for this special, but most important case, can be written 

Pp =^ %az^ 

r 

%az^ is called the polar moment of inertia and is designated by the letter J. 
The equation for torsion will finally be written 

Pp=^iJ (8) 

Let it be borne in mind that (4) holds only within the elastic limit, and for round shafts. 
This last limitation is sometimes overlooked or deliberately ignored. 

It can be easily demonstrated that if I^ and ly are moments of inertia referred to two 
rectangular axes with the origin at the centre of gravity of the section, J equals the con- 
stant sum of I^ and ly. Of course J can also be obtained by direct integration. 

10. Column Formulas. It is a familiar fact that slender or long columns fail by 
bending, and that a much smaller load is required to break a long column than to break 
a short one of the same material and cross-section. 

To allow for this in equation (1) a working value ofy^. is selected, diminished (from 
what would be allowable for a short block) to allow for the slenderness of the given 
column. 

Slenderness is expressed as the ratio of length to least width, or more commonly as 
ratio of length to a quantity r, known as the radius of gyration. This r is simply a 
length, such that Ar''^ = /. The proper rate of reduction of the allowable f to provide 
for slenderness is something for which the main reliance in actual design is upon experi- 
ments, guided somewhat by mathematical analysis. The results expressed algebraically 
constitute what is practically an empirical formula, called a column formula. 

Numerous column formulas are in use and the tendency is steadily towards the simplest 
of them. An example is given below. The condition of the end of the column, whether 
flat or hinged, is of much importance. Flat ended columns are prevalent in buildings, 
hinged ended are common in bridge-work and machinery. 

Columns whose - exceeds 145 or 150 are to be avoided, and for the best bridge work - is 

r ' to ,. 

not allowed to exceed 100 or 125. One reason for the large margin in the latter is the 
special danger from lateral shock through collisions, etc. 

For rolled sections, values of r will be found in the manufacturers' handbooks, and for 
rectangles it is to be taken 0.29 d, where d is shorter side of the rectangle. By substituting 
this value of r, slenderness appears as -5 in the formulas with the numerical coefficient modified 
accordingly. 

For this course the following column formulas are to be used : 



For mild steel, hinged ends, when - ^ 90, a/, = 17,100 — 73 - 



for - < 90, af, = 10,000 



12 



For mild steel, flat ends, when -. > 90, af, = 17,100 — 57 -, 

for^< 90, af, — 12,000. 
For white pine or spruce, flat ends, -, ^ 60, /. =: 625 — 0.15 ( t ) • 
For long leaf yellow pine, or white oak, flat ends, -= ^ 60, ./'. =i 900 — 0.20 I ^ 1 • 

" In applying the formulas to composite columns made up of several sticks bolted together at 
intervals, give to each stick the proj)ortionate share of the total load to be carried over that member, 
and then assume it stands alone and unsupported. This is the only safe rule. Even though they are 
firmly bolted, with packing blocks or wasliers notched into the sides, these grow loose in time, and do 
not resist lateral bending. They should never be assumed to act as one solid stick." * 

11. Allowable Stress. — Factor of Safety. All structural materials vary more or less 
in the properties of fundamental importance to the designer, viz. : w-^eight per cubic foot 
(specific gravity), modulus of elasticity, elastic limit, and ultimate strength under the 
various stresses. It is, moreover, impracticable, in the case of some materials, to set 
any definite value for some or even any of these quantities. Nevertheless certain unit- 
stresses have been arrived at through experience as "permissible" or "allowable" for 
the common structures subject to the usual loads. By "permissible" or "allowable" is 
meant having proper regard both to economy and to strength and durability. Such unit- 
stresses are also called working-stresses. They always provide in a structure a large mar- 
gin of strength against inevitable imperfections in workmanship, large variations in actual 
strength of the material, vibi'ations and unforeseen stresses and stresses impossible to 
estimate. This margin is expressed in the ratio of the load necessary to make the 
stresses rise to the ultimate strength, to the load actually anticipated ; that is, the ratio 
of the breaking load to the maximum load expected. This ratio is called the factor of 
safety, and varies from 3 or 4 to 10, 20, and more for various materials and stresses, 
and for various kinds of structures. 

Of course, as has been pointed out, the elastic limit is really a point of serious danger, 
and for the most uniform materials, like steel, is the logical though not the actual ultimate 
strength, and the factor of safety is coming to some extent to be stated with reference to 
elastic limit as a more scientific practice, involving of course lower nominal values for the 
factor of safety. 

12. Working Stresses for Timber as affected by Moisture, An element of great 
importance as regards the strength of timber is the amount of moisture it contains either 
as sap not yet seasoned out, or as water absorbed after seasoning. 

Bulletin No. 12, U.S. Department of Agriculture, Division of Forestry, says: — 

" Since the sti*ength of timber varies very greatly with the moisture contents, the economical 
designing of such structures will necessitate their being separated into groups according to tlie 
maximum moisture contents in use." 

* Professor J. B. Johnson, Mod. Framed Structures, p. 353. 



13 

This it proceeds to do as follows : — 

"Class A (moisture contents, 18 per cent.). — Structures fully exposed to the weather, such as 
railway trestles, uncovered bridges, etc. 

"Class B (moisture contents, 15 per cent.). — Structures under roof but without side shelter, 
fully exposed to outside aii-, but protected from rain, such as roof trusses for open shops and sheds, 
covered bridges over streams, etc. 

" Class C (moisture contents, 12 per cent.). — Structures in buildings unheated, but more or less 
protected from outside air, such as roof trusses or barns, enclosed shops and sheds, etc. 

" Class D (moisture contents, 10 per cent.). — Structures in buildings at all times protected from 
outside air, heated in the winter, such as roof trusses in houses, halls, churches, etc. 

" For long-leaf pine add to all the values given in the tables, except those for moduli of elasticity, 
tension and shearing, for Class B, 15 per cent.. Class C, 40 percent., and for Class D, 55 per cent. 
For the other woods make these percentages 8, 18, and 25 respectively." 

The working stresses for timber given on p. are taken from reputable practice for 
railway bridges and trestles. No equally well established values in building Avork exist. 
It would be not unreasonable to take the vakxes here given as values to which to apply 
the percentage corrections of the Division of Forestry, calling these "Class D," though 
the actual Class D recommendations of the Forestry Division vary somewhat from these. 

Among the many books bearing on the subject-matter of the course in addition to 

those mentioned above and the steel manufacturers' hand-books may be mentioned : — 

Greene's Structural Mechanics. 

Merriman's Mechanics of Mater i^ds. 

Merriman's Strength of Materials. 

Johnson's Materials of Construction. 

Howe's Simple Roof Trusses in Wood and Steel. 

Trautwine's Engineer's Pocket-Book, 17th ed. 

Kent's Mechanical Engineer's Pocket-Book. 

PROPERTIES OF TIMBER. 



Approx. 

Safety 
Factor. 



White Pine and 
Spruce, 



Wliite Oak. 



Long-leaf 
Pine. 



Weight per cu. ft. in lbs. (12 per cent, moisture) . . 
Apparent Elastic Limit (transverse) in lbs. per sq. in. 
Allowable Tension in lbs. per sq. in. 

with grain 

across grain 

Allowable Compression in lbs. per sq. in. 

end bearing 

columns <^15 diams 

across grain 

Allowable Shear, etc., in lbs. per sq. in. 

with grain 

across grain 

Allowable in Flexure, "Fibre stress" in lbs. per sq. in. 
Value of Modulus of Elasticity in lbs. per sq. in. . . 



10 
10 

5 
5 



24 + 
6,400 

700 
50 

1,100 

700 
200 

100 

500 

700 

1,000,000 



50 + 
9,600 

1,000 
200 

1,400 
900 
500 

200 

1,000 

1,000 

1,100,000 



.88 + 
10,000 

1,200 
60 

1,100 

700 
200 

150 

1,250 

1,200 

1,700,000 



14 



propp:rties of masonry. 



Brick Masouiv. 



First Class 
Stone Masonry. 



Stone Concrete. 



Granite Blocks. 



Weight per cu. ft. in lbs. 

Allowable Compression in lbs. per sq. in. 
Ultimate Flexure, in lbs. per sq. in. . 



100 to 140 
70 to 200 
40 to 80 



165 
250 to 400 



140 

70 to 210 

200 to 600 



170 



1,500? 



PROPERTIES OF IRON AND STEEL. 



Cast Iron. 



Wrought Iron. 



Steel. 



Weight per cu. ft. in lbs 

Elastic Limit in Tension in lbs. per 8(j. in 

Modulus of Elasticity in lbs. per sq. in 

Allowable Tension (in buildings) in lbs. per sq. in. 

Plates and Shai)es 

Eye-bars and Rods 

Allowable Tension (in railway bridges) in lbs. per sq. in. 

Plates and Shapes 

Eye-bars and Rods 

Allowable Compression in lbs. per sq. in. 

Short blocks (buildings) 

i > 90 

Allowable Shear, etc., in lbs. per sq. in. 

Pins and Rivets (buildings) 

Web Plates 

Allowable in Flexure, "Fibre stress" (buildings) 

in lbs. per sq. in 

Allowable Bearing, Rivets (buildings) in lbs. per sq. in. 



450 
15,000,000? 



480 

25,000 

25,000,000 

12,000 
15,000 



10,000? 



490 

30,000 

30,000,000 

15,000 
18,000 

9,000? 
8,100? 

12,000 
17,100 -57j 

9,000* 
7,000 

16,000 

18,000 



* For rivet steel. 



GENERAL INSTEUCTIONS REGARDING PROBLEMS 



The problems stated on the sheets bound herewith constitute a large part of the 
required work of the course, but they are to be understood as exercises, — as means of 
acquiring a knowledge of the principles of the subject through practice in their applica- 
tion. Each member of the class is expected to reflect with care after completing a prob- 
lem, and to be perfectly sure he has a clear understanding of the principle involved, 
or, at least, to know in just what particulars he needs to ask farther assistance. Whether 
this has been done and the knowledge really attained will be determined by the results 
of the written examinations, and frequent written tests. 

It is expected that help in doing the problems will be given freely by the teachers 
in the course (and moderate amounts by others, perhaps), and hence presenting solutions 
of the problems without satisfactory records in tests and examinations cannot constitute a 
sufficient ground for a pass marh in the course. On the other hand, tardy or otherwise 
unsatisfactory presentation of solutions of the problems may constitute failure to satisfy 
the requirements of the course regardless of records in tests and examinations. 

Attention is called to the fact that only those passing this course satisfactorily are 
eligible for admittance to the succeeding courses, Engineering 5c and 7a. 



1 A street railway steel wire cable ft. long and subjected to an 

avera<re tension of 12,000 lbs. per sq. in. How much is it lengthened thereby, assuming 
that i°ts modulus of elasticity is the same as that of a solid rod of the same material? 



2. The elastic limit of medium steel is about 35,000 lbs. per sq. in. How many 
inches would a tension rod of such steel long lengthen before suffer- 
ing permanent injury? How much would this lengthening be for a bar 8 in. long? 

Show the latter full size — one-quarter of an inch in diameter — in the original length 



and also in the length at the elastic limit. 



3. A medium steel rod is prevented from shortening while its temperature is 

YQ^^cla degrees F. Will the stress be beyond its elastic limit? 

Take the coefficient of expansion at 0.0000065. 



4. It is proposed to lift a weight of lbs. by cooling a steel rod 

15 ft. long and in diameter. Taking the coefficient of expansion at 

0.0000065, how many degrees must the temperature of the rod be reduced fairly to lift 



the weight off the ground ? 



5. Determine the proper diameter for a steel tension rod for a single king-post 
truss 32 ft. span, and 4 ft. deep. The truss is to carry a uniformly distributed load of 

lbs. per lineal foot. The segments of the top chord are to be con 

sidered to act like two simple beams as I'egards transmission of loads to the supports and 
the king-post. 

Assume (a) the rod to be made with upset screw ends ; 

(b) without upset ends, allowing 20 per cent, excess of cross-section at the 
root of the threads beyond the otherwise minimum requirement. 



O. A theatre gallery 40 ft. wide is supported at the rear edge by a series of girders, 

and at the front by round steel hangers in. in diameter at intervals of 

ft. and placed ft. back from the front edge. 

The gallery is subject to a total load of 200 lbs. per sq. ft. Are the hangers strong 
enough? If not, what should be their diameter? 



*T. (a) What load would probably be required to crush a cast-iron post 3 ft. long, 

in. in external diameter, and in. in internal? 

(5) What load would it safely support with a factor of safety of 10? 

Take uf^ = 90,000 lbs. per sq. in. 



S. A heavy dray body rests upon the axles at points very close to the hubs of 

the wheels. Diameter of the soft steel axle where it enters the hubs is 

inches. Taking a factor of safety of 10, what uniform load on the dray would the shear- 
ing strength of these axles justify, supposing the axles so placed as to divide the load 
equally. 

Take ufg at 50,000 lbs. per sq. in. 



O. A wooden triangular roof truss whose rise is of the span and 

whose bottom chord is horizontal, is to carry a uniform load of fiOO lbs. per lineal foot. 
The span of the roof is 60 ft. The top chord at its lower ends is to be notched into the 
top of the bottom chord. Width of both chords, 8 in. Wood, white oak. 

How deep should this notch be and how far, at least, should it l)e from the end of 
the bottom chord ? 

Joint to be of the type given. Take af(. = 500 + 900 sin a. Supposing each stick 8 in. 
square, draw a side elevation of the joint to scale, giving the calculated dimensions. 



10. I. Design a spruce cantilevei* to be 10 feet long and (a) to support a 

lb. safe hung from the end ; (b) to be one of the floor joists 5 ft. on centres of a balcony 
10 ft. wide, supported only at the wall, and to carry, including its own weight, a distrib- 
uted load amounting per sq. ft. to 2 per cent, of the weight of the safe. 

11. Design a spruce beam of 10 ft. in span (a) to carry this safe hung from its centre ; 
(b) to be one of a series of floor joists to carry same load per square ft. as in I. b. 
above. 

III. Determine and state the size of the steel beam to be selected for each of these 
four cases. State not only the beam with the least excess of — , but a preferable alter- 
native, should any such exist. 

— for rectangle = — ;- . Assume b = - . 

c => 2 

Enter data for Problem 11. 



11. Determine the cost of each of the eight beams of Problem 10, allowing 2 ft. 
entrance into the wall for the cantilevers, and taking the over-all length of the simple 
beams at 12 ft. 

Take, as the market price of spruce, |20 per thousand ft. B. M., and of steel, $2.50 
per hundred lbs. 



13. A portion of the fourth floor of a steel-frame office building is made up of 
rectangular bays 16 ft. 8 in. by 18 ft. with a column at each corner of each rectangle. 
Two joists and two girders are supported directly by each column, and two additional 
joists divide each bay into three rectangles 16 ft. 8 in. by 6 ft. 

(a) Calculate sizes for each of these beams, assuming the floor load (live and dead 
combined) to be lbs. per square foot. Suppose, though it is in some 
places not the practice, that the girders are to be figured for full live load. 

(b) Suppose the head-room available requires the girders to be n)ade of pairs of 
beams as shallow as possible, select proper sizes for these beams, sketch and dimension 
section of the girder, and make out a bill of bolts and separators with weights (p. 42). 

Separators should be placed near each end of the girder, near points of concentrated 
loading, and at distances apart not greater than twenty times the width of the smaller 
beam flange in order to prevent buckling. 

Writer data for Problem 31. 



13. A white pine stick 12 in. square and 24 ft. long is floating in a pond. It carries 
ll)s. at its centre. What is the stress on the extreme fibres, and 



which side is in compression and which in tension? 



14. Determine the fibre-stress due to its own weight in each of the following beams, 
eaoh supposed to be supported at both ends : 

(a) spruce 4 in. X 8 in long. 

(b) spruce " of ten times the length of (a) 

(c) 15 in. -50 lb. steel I of>the same length as (a) 
(<Z) 15 in. -50 lb. steel I of the same length as (6) 

Mention anything which may possibly invalidate the scientific correctness of the results 
of (c) and (d). 

Writer data for Problem 35. 



ISJ. Design a granite footing for a brick wall which supports, including its own 
weight, semi-tons per lineal foot, assuming the pressure uniformly dis- 
tributed under each layer. 

Allowable pressure on the foundation bed 7,000 lbs. per square foot. Allowable 
"fibre-stress" on granite 150 lbs. Thickness of courses of granite 12 in. What is the 
probable factor of safety here? What should be the thickness of the foot of the brick 
wall, allowing 100 lbs. per sq. in. on brick masonry? 

Suggestions. Determine the total width of the base of the footing, then the permissi- 
ble offsets (lengths of cantilever under a distributed load) and sketch to scale, using 
1 in. = 5 or 10 ft. Deduce and use a general formula. 



16. (a) What is the weight of a wrought iron rod 1 in. square, 1 yd. long? 

(b) From the result of (a) state at once the weight per ft. of a wrought iron bar of a 
section containing sq. inches. 

(c) What would be the sectional area of a wrought iron rail weighing lbs. 

per yard? 

(d) Steel is usually assumed to weigh 2 per cent, more than wrought iron ; what would 
be the weight per ft. of the bar of (b) if the material had been steel? 

(e) Compare the results of (b) and (d) with the Cambria data for one of 



UT". (a) Determine approximately I for a 6 in. X 12 in. rectangle for an axis through 
the centre of gravity parallel to the short sides, using arithmetic alone. 

Make four difterent approximations, first taking the surface as a whole and then suc- 
cessively as two, four, and eight strips of uniform width. 

(b) Determine by integration 1 for a rectangle of breadth, b, and depth, d. 
From the result determine numerically the exact value of the I of (a). 

(c) Determine by integration I for a circle of radius r. Polar coordinates will be 
most convenient. 

Note that I is merely a name given to a geometrical property of a figure which in 

connection with c (— ^ is a measure of the effectiveness of the distribution of the material 
to resist bending. 



IS. The area of any surface is A ; its I referred to XqXo (any axis through its 
centre of gravity) is Iq ; X'X' is an axis parallel to XqXo and distant from it by h; 
I' is the I of the surface referred to X'X'. 

(a) Prove that I' = Iq + Ah^. 

Be sure that it is clear why s ay (for C. G. axis) always equals zero. 

(&) Does it matter on which side of XqXo X'X' is taken? Wh_y? 

(c) Would neglecting Iq systematically lead to too large or too small a value for I, — 
i.e. to a value larger or smaller than the correct one? 

Note that the errors due to the approximations in Problem 17 (a) above arose simply 
from the neglect of the Iq's of the strips. As the strips grew shallower, the Iq's becoming 
rapidly less, this error rapidly diminished. 



19. Check values given in the hand-book for a , neglecting the 

rounding of corners, for 

(a) Area of section ; (^) ^ 5 

(b) Weight per foot ; (/) r' ; 

(c) I; (g) s; 

(d) I' (Take the given location of C. G. in case of channels) ; (h) Find the value of s'. 
Begin by making a clear sketch to a liberal scale, dimensioned fully. See pp. 2-11 for 

dimensions in addition to those given in the table of properties. 

State each result clearly with the hand-book result entered beside it in parentheses for 
comparison. 

Enter data for Problems 41, 56, 20. 



;30. Determine the maximum load which the section of Problem 19 could carry 
distributed or concentrated at mid-span regardless of its length. Hand-book, pp. 66-69. 
Use the formula of p. 67 and state result with value from table of next page beside it 
for comparison. 

Determine also the minimum span for which the beam could be trusted with the full 
uniform load suggested by the bending strength. 



Ql. A group of (i three-quarter in. rivets in single shear was sheared off by a pull 

of 00 lbs. What was the ultimate shearing strength of the rivets per 

square inch? 



22. A truss consisting of a series of eleven equilateral triangles whose sides are 

each 25 ft. long has a single concentrated load of semi-tons at the 

second top chord joint from the right. The bottom chord is made up of i)airs of eye- 
bars. Determine a suitable selection of such liars, both in number and size, for the third 
bay from the left. Take af at 12,000 lbs. per sq. in. since the truss is subject to vibra- 
tions. Neolect the weii>ht of the truss itself. Sketch one-half of one of the bars, oivino- 
all dimensions (p. 2i)3), to a scale 1| in. to the foot. 

Ignoring the diagonals and their efl'ect upon the pin, determine the size of the pin, 
taking the allowable fibre-stress at 18,000, crushing stress at 15,000 and shear at 
9,000 lbs. per sq. in. Check the diameter required for bending resistance by aid of the 
hand-book (pp. 274-275). 



23. A boiler is to be formed of wrought-iron plates f in. in thickness united by- 
lap joints with rivets | in. in diameter and with 1| in. in pitch. What is the ultimate 
efficiency of the joint, making allowance of -^^ in. in the diameter of the holes to offset 
injury from punching and an additional sixteenth clearance for the rivets in the holes. 

If the boiler is in. in diameter and the steam pressure is 

lbs. per sq. in., what is the factor of safety? 

Take uf for shear 50,000, for bearing 8,000, for tension 65,000. 



34. Check the value of "x" for ., and also of "c" of p. 201. 

(See also pp. 48-50.) 

Determine also the value of "c," ignoring the I' of the section. 



3S». (a) Design a medium steel column to be 24 ft. long to support 



semi-tons, ends of column fiat; - not to exceed 125. State also the safe load for this 



r 



column selected. 

Suggestion: Seek among I beams, channels, Z's, T's, L's, pairs of angles (pp. 189-193), 
two pairs of angles latticed or a pair of channels latticed, for something offering such a 
suitable value for r. Record your success with each of these shapes with weight per ft. 
of resulting column. 

(6) Solve a second time for a load four times that given above. 

Transfer data and result to next problem-sheet for reference. 

Enter data also for Problem 29, a and b. 



2G. Re-design the column of Problem 25, using the smaller of the two loads, allowing 

a value of -: as high as 250, if any metal can thereby be saved. Take the same steps as 

Problem 25, recording results of each trial. State also the safe load for this column 
selected. 



S'T'. What would be rated as the safe load for a in. X in. 

white oak column ft. long? Square ends. 



2^. Check the hand-book values foi- Z-bar column given (p. 210) as safe for 

tons for a length of 12 ft., for (a) cross-section; (b) weight per ft.; 

(c) least r, and (d) safe total load for the limiting length given. 

(e) What would such a column cost, if 12 ft. long and 2^ cents per lb. for steel were 
quoted ? 

For (d) use both Gordon's formula and the straight line formula. 

Record your results and the hand-book figures in form for comparison. 

If the given load exceeds the limiting load of the tables, halve it. Use hand-book 
values of Iq so far as possible. 



2&. (a) What would be least width ovei' all of a column of the channels of Prob- 
lem 25 above, if the channels could be placed with the flanges inwards? 

(b) How much over-all width could thus be saved? 

(c) What would hinder the channels from being thus set face to face? 



30. Deduce the fundamental formula for dealing with the simplest case of pure 
bending. State clearly the fundamental hypothesis. 
When only is the formula strictly applicable? 
Why must M = S^? 



31. Select columns (pp. 210-243) for the third story (height of story being 13 ft. 
6 in.) of the building of Problem 12, supposing the load on each of them above fourth 

floor to be tons, and supposing (as is not usually done) that the columns 

are figured for full live and dead load. 

Sketch free-hand a cross-section of the column chosen showing sizes of steel shapes 
used. 



3S- A beam is fixed rigidly in a wall at one end and projects 

inches. Forces act on the overhanging portion as follows: (200 lbs., 30°, 0,0), ( 

lbs., 240°, 20,0), (1,000 lbs. 150°, 36,0), (1,200 lbs., 315°, 50,0). Origin at the centre of 

the free end, and axis of X coincident with axis of beam. 

Determine stresses (normal stress, shear, and bending moment) fully at mid-span and at 
the wall, both graphically and algebraically, and record results for comparison. 

For algebraic work, replace each force by horizontal and vertical components. 

For graphic work, use forces as given. Scale 1 in. =i 10 ft. and 1,000 lbs. 



33. Check A'alues of 

(a) To ; (b) i\ ; (c) r^ ; (d) 1-3 which are given on pp. 181»-193 of the 

hand-book for two angles legs together, taking such 

data as may be convenient from other tables. 

Use data from pp. 168-183 as far as possible. 



34. A wrought-iron anchor-rod for a tower is likely to be subjected to an initial 
tension of 12,000 lbs. before being left bj'^ the workmen. The wind pressure stress in the 
rod is lbs. 

Determine the diameter of rod (a) if made with upset screw ends; (b) if not so made. 



3S. Find the probable deflection in each case of Problem 14. Sketch the size and 
length of each beam. 

See hand-book, pp. 139-140. For further relating to the treatment of deflection of 
beams, those interested are referred to Merriman's Mechanics of Materials, pp. 70 et seq. 



36. It is proposed to break a round rivet-steel bar 2 ft. 6 in. long, weighing 
lbs. (a) How large a force must be provided? 

(b) Sketch the outline (to large scale) of this bar as it would look when about to 
break, showing its probable total length and least diameter. 

(c) About what would you expect to find the elastic limit? 

For further data see hand-book, p. 309. Reduction of ai'ea may be expected to be 
about 60 per cent. 



3*7^. It is proposed to make a steel street-railway line without joints in the rails, 
relying upon the grip from the ballast, ties, etc., to prevent expansions and contractions 
from tempex-ature. Supposing the rails to be laid at 40° F., and that it is expected they 
may reach a temperature of S'0° F. on the average throughout their section, and assuming 
a coefficient of expansion of 0.0000065, state how much must be the frictional grip per 
lineal foot of rail, and what would be the resulting stress per sq. in. of the rail. 



3S. A double-riveted lap-joint is to be made with plates | in. thick, and | in. 
rivets. Determine tlie pitch so that the strength of the plate shall equal the shearing 
strength of the I'ivets, and calculate the efficiency of the joint. 

Take ut; = 60,000, uf, = 50,000, and uf; = 80,000. 



30. It is proposed to use a 10 in. -25 lb. I to carry a uniformly distributed load 
of 5.9 tons, with supports 22 ft. apart. A plastered ceiling is to be used below the beams. 
Question is raised whether beams will be stiff enough to prevent ceiling from cracking. 
Is the point well taken, a deflection of ^L- the span being the maximum allowable over 
plastered ceiling? See hand-book, pp. 73, 139, 140, and Merriman's Mechanics of 
Materials, p. 76. 



40- A cylindrical steel boiler ft. in. in 

diameter is to be subject to 150 lbs. per sq. in. internal pressure. A tensile stress on 
the steel above 5,000 lbs. per sq. in. is not considered safe. Determine minimum permis- 
sible thickness of steel, making no allowance for the weakening by riveting. 

Is the boiler more likely to open a lengthwise or a circumferential joint? 



41. (a) Deduce an expression, C, in terms of familiar letters which when divided 
by the length of a given beam in ft. will give the total distributed load in lbs. which the 
beam would safely carry. 

(b) Find the numerical value of C for shape of Problem 19 if the allowable fibre-stress 
be taken at 16,000 lbs. per sq. in., at 12,500 lbs. per sq. in., and at 10,000 lbs. per sq. in. 

(c) Record hand-book values for these C's so far as possible. 
Enter data in Problem 42. 

Observe that in this problem, together with 19 and 24, there has been checked an 
item in every column of properties of beams and channels. Properties of other sections 
are of course found in a similar vi^ay. 



43. By use of C with 12,500 fibre-stress state the safe load for the shape of Prob- 
lem 41 in each of the following cases : — 

(a) load concentrated at mid-span, span 20 ft. 

(b) as a cantilever under disti'ibuted load, span 20 ft. 

(c) as a cantilever with load concenti'ated at end, span 16 ft. 8 in. 

(d) as a cantilever with load concentrated at mid-span, span 16 ft. 8 in. 



43. A mill construction building has five stories averaging 12 ft. in height, and a 
basement whose floor is 13 ft. below the top of finished first floor. Column bases are two 
feet above the basement floor. Columns are 12 X 12 yellow pine. Roof is supported 
entirely by the walls. 

Supposing each column responsible for a floor space ft. X ft. 

and that all floors may be called upon to support a load of lbs. per square 

ft. at once, 

(a) how much settlement from compression of the columns should be expected at 
the fifth floor? 

(6) how much at the second floor? 



44. A bowstring truss, whose web members are to be diagonals and verticals, is 

to be of G equal panels of 10 ft. each. Concentrated loads of 120, 36, and 

semi-tons are to be supported from the lower chord joints 20, 30, and 50 ft. from left end. 

Design the second vertical from the left, and show detail of the end connection to scale. 
Assume gusset plates | in. thick, and allowable shear and bearing on rivets 7,500 and 
15,000 lbs. per sq. in. respectively; | in. rivets. 

Take af^ = 12,000. 



4^- A rod 2 in. in diameter broke under tensile stress of lbs. 

What was its ultimate stress i)er sq. in. ? 



4C5. A yellow pine beam 6 in. X 8 in., 12 feet between supports, failed in flexure 

under centre load of lbs. How much was the modulus of rupture of 

the wood? 



47^. The rim of a cast-iron fly-wheel is subjected to tension of 5,000 lbs. per sq. in. 
from centrifugal force. How much is its diameter increased if its original diameter was 
ft in.? 



4©. Beams ft. in. c. to c, ft. span. 

(a) What load per sq. ft., including its own weight, will beams carry if they are (a) 15 in. 
42 lb. I, (b) 4 in. X 14 in. yellow pine, (c) 10 in. X 12 in. spruce, and they are covered 
with in. yellow pine planking? 

(b) For what net load per sq. ft. would floor l)e rated safe in each case? 



49- Determine the polar moment of inertia of a circle of radius r, by integration. 



^O. (a) Find the diameter of a solid steel shaft to transmit H. P. 

at 112 revolutions per minute, the maximnm twisting moment being expected to be 1.3 
the mean, and stress allowed 10,000 lbs. per sq. in. 

(6) Find the size of a hollow shaft to replace the preceding, diameter of whole | the 
outside diameter. Estimate the saving in weight in 40 ft. of shafting. 



SI. A X I in. steel plate is to be spliced with a double cover 

butt joint in such a way as to make the joints as nearly as possible as strong as the 
body of the plate. 

Sketch the joint carefully, showing the proper number and arrangement of the rivets, 
and state the efficiency of the joint. 

Count a rivet hole as destroying steel for a space ^ in. more than diameter of rivet. 



S»S. Bay window cantilevers of which a sketch will be given. Design 
method of supporting them at the back. Steel frame building to be designed for total 
floor load of lbs. per sq. ft. 

Allowable tension 12,000, shear on rivets 10,000, bearing 20,000. 

Suggestion : Determine max. moment ; the pull on strap, number of rivets required, 
size of strap. How provide for shear? See hand-book, pp. 270, 273, 43, 50. 



S3. Design concrete and steel-beam footing for column carrying 

tons. Ground regarded safe for 5,000 lbs. per sq. ft. Fibi'e-stress on beams 20,000 lbs. 
State size of cast-iron shoe. Neglect flexural strength of concrete. Compare hand-book, 
p. 263, and Carnegie, i)p. 1(37, 168 ; Freitag's Architectural Engineering, pp. 171-190. 

\ It will be found convenient first to deduce general formulas in which your varying 
data can be substituted. 



S4. A uniformly distributed load of tons is to be carried in span 

of 20 ft. Will any single beam safely do it? What can be used (pp. 247-262)? Sketch 
cross-section of the structure selected. 



S»S. In a certain floor to carry lbs. per sq. ft. joists to span 20 ft. 

are to be spaced 6 ft. apart. Columns 18 ft. apart, (a) Show two natural ways for set- 
ting the joists with reference to columns, (b) Compare the weights of the girders required 
in the two cases. 



5G. Design square wall plate for brick work good for 150 lbs. per sq. in. for section 
of Problem 19, carrying its full safe load (af = 16,000) over a span 12 times its depth. 
Fibre-stress in plate 16,000 lbs, per sq. in. Cf. pp. 46, 47. 



ST. A series of segmental brick arches of 8 ft. span are cai-ried by steel joists of 
20 ft. span. Rise one-eighth of the span. Arch 12 in. thick. The total load for the 
floor may be taken at 300 lbs. per sq. ft. 

Design the outside beam of the set and state the spacing you would set for | in. 
tie-rods. 



SS. A simple triangular truss of 20 ft. span and 6 ft. rise carries lbs. 

at its apex. 

Determine proper sizes for the three sticks and design joint at foot of rafter. All 
yellow pine. Secure at ridge against buckling. 

Give all computations and draw to scale sketch of result. 



so. Determine the most economical arrangement of beams for the floor system of 
a building 90 ft. X 40 ft., to carry same total load as in Problem 52. Fireproof floor to be 
used will permit joists to be spaced 8 ft, on centres. 



C30. A long-leaf yellow pine beam 12 in. X 14 in. is on supports 10 ft. apart and 
about to be tested to destruction by centre load. How would it probably fail? 
See foot of page of these notes. 



61. A masonry dam whose total height is ft. Its up-stream face 

is vertical for a distance of one-third of its height from the top, and thence has a batter 

of one horizontal to vertical. The down-stream face has a uniform 

batter of one horizontal to four vertical. Dam 20 ft. wide on top. 

Required lines of pressure, (o) when reservoir is empty : (6) when reservoir is full. 

Weight of masonry to be taken at lOO lbs., and of water 62.5 lbs. per cu. ft. respectively. 

Required the unit pressure on the foundation at heel and toe of dam. 

Statical work to be done graphically. 



C53. Investigation of Masonry Arch. (See Baker's Masonry Construction, §§ 695, 
679, 689, and 700.) 

A semi-circular arch has a span of ft. Uniform depth of voussoirs 

one-twelfth of the span. No masonry backing. Loaded with earth brought to a level grade 
6 ft. above crown of the arch. 

Taking weight of earth at 100 lbs. and of masonry 150 lbs. per en. ft. respectively, 
investigate stability of the arch. Assume that earth exerts no horizontal pressure. 
(Sheffler's Theory.) 



63. Design of a Warehouse Roof. 

Type of truss triangular, also called English. 

Span 60 ft. Rise of span. 

Eight panels of uniform width. Trusses 8 ft. apart on centres. Verticals of steel ; 
rest Georgia pine. 

Take loads as follows, consulting Carnegie, Merriman's Roofs and Bridges, Parts 1 
and III, or other reference-book for details not given here. 

Permanent load — slate covering on 1^ in. boards not finished below. 

Jack rafters and purlins — allow 2^ lbs. per sq. ft. of plan. 

Weight of truss itself to be estimated from 



W 



= - f 1 + / \ Merriman R. & B., Part I, p. 3, 

2 \ lOy 



in which W = approximate weight of wooden truss in lbs., a = distance between trusses, 
I ^ span of truss a and I, both in feet. Consider permanent load all carried by top chord. 

Snow — take 20 lbs. per sq. ft. of plan. 

Wind — provide for 20 lbs. per sq. ft. normal to roof slope, also, for designing anchoi- 
age only, 10 in. per sq. ft. Upward normal to under slope of rcof surface over and above 
the weisfht of roof itself. Each wall to take half the wind thrust. 

Determine maximum and minimum stresses in all the members graphically, checking 
each stress diagram by algebraic application of method of sections. 

Two stress diagrams needed — one for normal loads, one for vertical. 

Use tabulation to be given out for combining stresses. 



04. Determine the height of a prismatic brick tower that might be expected to 
crush under its own weight. Weight of brick masonry 120 lbs. per cu. ft. Take ulti- 
mate strength at 1,500 lbs. per sq. in. 



OS. A derrick mast and boom are each 12 in. X 12 in. long-leaf yellow pine. The 

mast is 50 ft. high, the boom ft. in. long. Gruys are 

never to be inclined more than 30° to the horizontal. The weakening of the jib due to 
deflection under its own weight would be considerable. 

Taking factor of safety of 10 as suitable, work out the actual working capacity of the 
derrick justified by these data. 



MAR 17 1903 



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